Question

x^2/3 + 4x^1/3 - 5 = 0

Answer 1

quberoot(x^2) + quberoot(4x) - 5= 0

Answer 2

you can solve this as a quadratic equation since the ratio of the exponents of the first two terms is 2:1\[\left( x^{1/3} \right)^2+4\left( x^{1/3} \right)^1-5=0\]

Answer 3

solve as usual by factoring by solve for x^(1/3) instead of the usual x...

Answer 4

Yes

Answer 5

If you can please

Answer 6

ok

Answer 7

\[(x^{1/3}+5)(x^{1/3}+1)=0\]\[x^{1/3}+5=0 \cup x^{1/3}-1=0\]\[x^{1/3}=-5 \cup x^{1/3}=1\]\[x=(-5)^3=-125 \cup x=1^3=1\]Solution set is {-125, 1}

Answer 8

Thanks

Answer 9

happy to do it